/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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NO

After renaming modulo the bijection { a ↦ 0, b ↦ 1, c ↦ 2 },
it remains to prove termination of the 4-rule system
{ 0 ⟶ ,
  0 0 0 ⟶ 1 ,
  1 2 ⟶ 2 0 2 1 ,
  2 ⟶ }

The system was reversed.

After renaming modulo the bijection { 0 ↦ 0, 1 ↦ 1, 2 ↦ 2 },
it remains to prove termination of the 4-rule system
{ 0 ⟶ ,
  0 0 0 ⟶ 1 ,
  2 1 ⟶ 1 2 0 2 ,
  2 ⟶ }

Loop of length 12 starting with a string of length 5
using right expansion and the encoding { 0 ↦ a, 1 ↦ b, ... }:

.cb.bba
	rule cb-> bcac at position 0
.bcac.bba
	rule a->  at position 2
.bcc.bba
	rule cb-> bcac at position 2
.bcbcac.ba
	rule a->  at position 4
.bcbcc.ba
	rule cb-> bcac at position 4
.bcbcbcac.a
	rule cb-> bcac at position 3
.bcbbcaccac.a
	rule c->  at position 6
.bcbbcacac.a
	rule c->  at position 6
.bcbbcaac.a
	rule c->  at position 7
.bcbbcaa.a
	rule aaa-> b at position 5
.bcbbcb.
	rule cb-> bcac at position 4
.bcbbbcac.
	rule c->  at position 5
.bcbbbac.